Download Physics_Test_01

Test 1
Core short answer questions: 50 marks
Suggested time: 60270 minutes
Core short answer questions
Specific instructions to students
• Answer all questions in the spaces provided.
• For all questions which require a numerical
answer you must show all working.
• You should take the value of g to be 10 ms22.
QUESTION 4
2 marks
As the tram begins to accelerate, the standing passengers
in the tram feel themselves ‘thrown’ backwards slightly.
Explain this phenomenon in terms of Newton’s first law
of motion.
Answer:
This phenomenon is due to the passengers’ inertia.
The passengers were travelling at constant velocity.
Relative to the accelerating tram, the passengers
A tram of mass 10 000 kg is travelling east along a street at
constant velocity. There is a total frictional force of 1500 N
acting on the tram.
continue to travel at constant velocity due to their
inertia which is slower than the tram which is accelerating,
thus they move backwards relative to the tram.
QUESTION 1
2 marks
According to Newton’s first law of motion, what must be
the driving force supplied by the tram’s motor?
Fnet 5 0
Fdriving 5 Ffriction 5 1500 N
Answer:
1500 N
QUESTION 5
2 marks
At one instant the tram is travelling east with an
instantaneous speed of 7.0 ms21. The tram slows down
at a constant rate and comes to a stop over a distance of
62 m. What is the magnitude of the tram’s deceleration?
u 5 7, v 5 0, s 5 62, a 5 ?
v 2 5 u 2 1 2 as
0 5 72 1 2 3 a 3 62
QUESTION 2
2 marks
Later, the tram is accelerating in an easterly direction, at a
rate of 0.25 ms22. What is the net force acting on the tram?
Fnet 5 ma
Fnet 5 10000 3 0.25 5 2500 N
49 5 2124 a
a 5 49 4 2124
a 5 20.395 ms22
Answer:
0.4 ms22
Answer:
2500 N
QUESTION 3
2 marks
If the total frictional force acting on the tram is still
1500 N, what is the driving force supplied by the tram’s
motor now?
Fnet 5 2500
Ffriction 5 1500
2500 5 Fdriving 2 1500
Fdriving 5 2500 1 1500 5 4000 N
QUESTION 6
2 marks
How much time will pass while the tram comes to a stop?
u 5 7, v 5 0, s 5 62, t 5 ?
s 5 ½ (u 1 v ) t
62 5 ½ (7 1 0) t
t 5 62 4 3.5
t 5 17.7 s
Answer:
18 s
Answer:
4000 N
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To demonstrate circular motion to her students, a Physics
teacher attached a 40 g rubber stopper to a length of
fishing line and spun it clockwise in a horizontal circle
when viewed from above.
At this instant the forces (ignoring friction) acting on the
rubber stopper would be:
P
QUESTION 7
2 marks
When the rubber stopper is at point P as shown in the
diagram above, the direction of the stopper’s instantaneous
velocity and acceleration when viewed from above
would be:
A
a
v
D
a
a
C
D
Answer:
A
The weight force acts vertically downwards, the
hence answer A.
a
C
B
tension in the fishing line acts along the fishing line,
B
v
A
v
v
QUESTION 9
2 marks
If the rubber stopper completes a horizontal circle of
radius 50 cm in 0.8 s, what is the average speed of the
rubbber stopper?
v 5 2r/T
v 5 2 3  3 0.5 4 0.8 5 3.927 ms21
Answer:
Answer:
B
The velocity is tangential to the motion at P, and the
acceleration is directed towards the centre of the
circle, hence answer B.
QUESTION 8
2 marks
At one instant in the motion, one of the students in the
class has the following side-on view of the rubber stopper
and length of fishing line.
e
in
gl
4 ms21
QUESTION 10
2 marks
In the above situation, what would be the size of the net
force acting on the rubber stopper?
Fnet 5 mv 2/r
Fnet 5 0.04 3 3.9272 4 0.5 5 1.234 N
Answer:
1.2 N
hin
fis
rubber stopper
2
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QUESTION 11
3 marks
In the previous situation, what would be the size of the
tension force in the fishing line?
W 5 mg
A
W 5 0.04 3 10 5 0.4 N
T5
_____________
​ (1.2342 1 0.42) ​5

QUESTION 14
2 marks
Which of the four alternatives below best represents the
acceleration–time graph for the vertical motion of the
rubber stopper as a projectile?
B
1.297 N
Fnet
W
T
C
D
Answer:
1.3 N
QUESTION 12
3 marks
At one instant during the above motion, the fishing line
snaps and the rubber stopper flies off horizontally as a
projectile. If the rubber stopper was 1.8 m above the floor
when the fishing line snapped, how long will it remain in
flight?
Answer:
B
Vertically, acceleration is always constant at 210 ms22
due to gravity, hence answer B.
Take downwards as positive
Analysing the vertical motion of the rubber stopper:
u 5 0, a 5 10, s 5 1.8, t 5 ?
s 5 ut 1 ½ at 2
_____________
t5
​ (1.8 3 2 4 10) ​5 0.6 s
Answer:
QUESTION 15
2 marks
Which of the four alternatives above best represents the
vertical net force2time graph for the motion of the rubber
stopper as a projectile?
Answer:
B
0.6 s
Net force 5 mass 3 acceleration, so the net
force–time graph has the same shape as the
QUESTION 13
2 marks
How far horizontally will the rubber stopper travel from
where the fishing line snapped until it hits the floor?
Analysing the horizontal motion of the rubber stopper:
u 5 3.927, a 5 0, t 5 0.6, s 5 ?
s 5 ut 1 ½ at2
acceleration2time graph, hence answer B.
A skier of mass 60 kg is travelling down an incline at an
angle of 16° to the horizontal at a constant speed of 4 ms21
in a straight line.
s 5 3.927 3 0.6 1 0 5 2.356 m
Answer:
16°
2.4 m
QUESTION 16
2 marks
In which direction is the net force acting on the skier?
A
B
C
D
zero net
force, no
direction
Answer:
D
The skier is travelling at constant speed in a straight
line and therefore has no acceleration and no net force.
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QUESTION 17
3 marks
Calculate the magnitude of the total frictional forces
acting on the skier.
Since Fnet 5 0, Fdown the slope 5 Ffriction
Fdown the slope 5 mg sin
QUESTION 21
2 marks
The thief, even though he was wearing a seatbelt, will
experience a different impulse to the car. Explain why
this is so.
Answer:
5 60 3 10 3 sin 16°
Impulse 5 change in momentum 5 mass 3 change
5 165 N
in velocity. The seatbelt means the thief has the same
Answer:
165 N
change in velocity as the car, but his mass is much
less than the mass of the car so his impulse is also
much less.
QUESTION 18
4 marks
At a later time on the same slope, the skier is accelerating
at a rate of 0.5 ms22 down the slope. Show that the
magnitude of the total frictional forces acting on the skier
is now 135 N. Include your working.
Fnet 5 ma
5 60 3 0.5
5 30 N
QUESTION 22
3 marks
Fortunately for him, the car thief survives the crash due
in part to the car’s airbag. Explain, using the principles of
physics, how this device reduces the impact force on the
driver in car crashes.
Answer:
Since impulse 5 force 3 time, an increase in the time
Fnet 5 Fdown the slope 2 Ffriction
of a collision causes a decrease in the impact force on
30 5 165 2 Ffriction
the driver for the same amount of impulse. An airbag
Ffriction 5 165 2 30 5 135 N
cushions the driver and causes them to come to a stop
Answer:
135 N
over a longer period of time than without the airbag,
hence reducing the impact force of the collision.
A car thief is about to have a very unpleasant accident. He
is driving a stolen Alfa of mass 1.1 3 103 kg at 92 kmh21
when he fails to take a bend in the road, causing him to
crash head-on into a telegraph pole.
QUESTION 19
2 marks
What is the magnitude of the car’s change in momentum
if the car comes to a stop after the collision?
p 5 mv
p 5 1.1 3 103 3 92 4 3.6 5 28111 kgms21
Answer:
2.8 3 104 kgms21
QUESTION 20
2 marks
What is the magnitude of the impulse on the car during
the collision?
I 5 p 5 28000 Ns
Answer:
2.8 3 104 Ns
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